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Is $f^*E$ (the pullback of a smooth bundle) an embedded submanifold of $N\times E$?

I know that it's well defined and at least immersed:

Let $\pi: E \longrightarrow M$ a smooth fiber bundle and $f:M \longrightarrow N$ a smooth function. The Poor's book constructs the pullback bundle as the subset $f^*E:=\{(p,x) \in N\times E|\; f(p)=\pi(x)\}$ with projection $(f^*\pi)(p,x)=p$ and "trivialization charts":

$\Psi:(f^*\pi)^{-1}[f^{-1}[U]] \longrightarrow f^{-1}[U] \times F$

$\Psi(p,x)=(p,pr_2(\psi(x)))$

With $\psi:\pi^{-1}[U] \longrightarrow U\times F$, a bundle chart of $\pi:E \longrightarrow M$

I've already proofed that there exist a (unique) smooth structure in $f^*E$ such that $f^*\pi:f^*E\longrightarrow N$ is a smooth fiber bundle with trivialization charts $(f^{-1}[U],\Psi)$. By calculus in these charts, the inclusion $i:f^*E \longrightarrow N\times E$ is an immersion, so $f^*E$ is an immersed submanifold of $N\times E$.

Is it an embedded submanifold? Why? Is there an easy way to show that the inclusion is an embedding?

Thanks


I found a post with the same question, the answer is interesting because it involves less structure, the minmum to apply transversality, but I think it has to exist a more easy way in the case. The Pullback Bundle is an Embedded Submanifold of its Parent Space

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  • $\begingroup$ No, you have things backwards. If $E$ is a bundle on $M$, then you need a smooth function $f\colon N\to M$ (not the other way around), and then $f^*E$ will be the bundle on $N$ whose fiber over $p$ is the fiber of $E$ over $f(p)$. $\endgroup$ – Ted Shifrin Dec 10 '19 at 22:39
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There is no secret. Let us start with definitions and some well-known facts.

Let $\phi : M_1 \to M_2$ be a smooth map between smooth manifolds $M_1, M_2$. It is called a (smooth) embedding iff $\phi(M_1)$ is a smooth submanifold of $M_2$ and $\phi : M_1 \to \phi(M_1)$ is a diffeomorphism. It is called an immersion if all derivatives $T_p \phi : T_p M_1 \to T_{\phi(p)} M_2$ are injective. Each immersions is locally an embedding (which means that each $p \in M_1$ has an open neighborhood $U$ such that $\phi\mid_U$ is an embedding). Immersions are in general no embeddings (even if they are injective). However, if an immersion is a topological embedding (which means that $\phi : M_1 \to \phi(M_1)$ is a homeomorphism), then $\phi$ is an embedding. See for example https://www.math.lsu.edu/~lawson/Chapter6.pdf.

Starting with the subspace $M_1 = f^*E$ of $M_2 = N \times E$, you have shown that the space $M_1$ can be endowed with a smooth structure which is uniquely determined by suitable requirements. You have moreover shown that the inclusion $i : M_1 \to M_2$ is an immersion. But $i$ is a topogical embedding by definition. Thus $f^*E = i(f^*E)$ is a smooth submanifold of $N \times E$.

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  • $\begingroup$ Thanks, now I understand. For me $M_1$ was a subset, I defined its topology as the unique such that $\Psi$ were bundle charts. I see that the functions $\Psi$ and $\Psi^{-1}$ are continous putting the subspace topology in $M_1$. So the previuos topology I've found is the subspace topology and the inclusion is a topological (and smooth) embedding. $\endgroup$ – alexpglez98 Dec 15 '19 at 18:22

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